Notice that this is a Riemann sum with infinitely many rectangles.

And a Riemann sum with infinitely many rectangles is the Definition of an Integral:

$\lim \limits_{x\rightarrow \infty }\displaystyle \sum_{n=1}^{n}\Delta xf(a+\Delta xi)=\int_{a}^{b}f(x)dx$

Can you rewrite this as an integral?

$\lim \limits_{n\rightarrow \infty }\frac{1}{n}=\lim \limits_{x\rightarrow 0}\Delta x=dx..............\text{ so we can substitute }\frac{1}{n}\text{ with }dx.$

$\lim \limits_{n\rightarrow \infty }dx\left [ \frac{1}{\left ( \frac{1}{n} \right )}+\frac{1}{\left ( \frac{2}{n} \right )}+...+\frac{1}{\left ( \frac{n}{n} \right )} \right ]$

The $dx$ represents the infinitely small width of each rectangle.
Now lets find the height of each rectangle.
Heights, of course, are represented by a function, $f\left(x\right)$.
But what is $f\left(x\right)$?

Since $x$ is a variable, we will let $x$ represent the part of the series that is changing:

$\frac{1}{n}+\frac{2}{n}+...+\frac{n}{n}.$

$\text{Thus the function must be }f(x)=\frac{1}{x}...............\text{ substitute.}$

$\text{This is beginning to look more like an integral: }\lim \limits_{n\rightarrow \infty }\frac{1}{x}dx$

We still need to find the bounds of the integral.

$\text{Return to the values of }x: \frac{1}{n}, \frac{2}{n}, \frac{3}{n}, ... , \frac{n}{n}.$

$\text{The lowest value of }x \text{ is } \frac{1}{n}. \text{ Since }\lim \limits_{n\rightarrow \infty }\frac{1}{n}=0,\text{ the lower bound is }0.$

$\text{The highest value of }x \text{ is } \frac{n}{n}. \text{ Since }\lim \limits_{n\rightarrow \infty }\frac{n}{n}=1,\text{ the upper bound is }1.$

$\text{Put it all together: }\int_{0}^{1}\frac{1}{x}dx$